Question: Simplify and expand the following expression: $ \dfrac{3q}{q - 5}-\dfrac{2q - 5}{3q - 4} $
Explanation: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(q - 5)(3q - 4)$ Multiply the first term by $\dfrac{3q - 4}{3q - 4}$ $ \begin{align*} \dfrac{3q}{q - 5} \times \dfrac{3q - 4}{3q - 4} & = \dfrac{(3q)(3q - 4)}{(q - 5)(3q - 4)} \\ & = \dfrac{9q^2 - 12q}{(q - 5)(3q - 4)}\end{align*} $ Multiply the second term by $\dfrac{q - 5}{q - 5}$ $ \begin{align*} \dfrac{2q - 5}{3q - 4} \times \dfrac{q - 5}{q - 5} & = \dfrac{(2q - 5)(q - 5)}{(3q - 4)(q - 5)} \\ & = \dfrac{2q^2 - 15q + 25}{(3q - 4)(q - 5)}\end{align*} $ Now we have: $ = \dfrac{9q^2 - 12q}{(q - 5)(3q - 4)} - \dfrac{2q^2 - 15q + 25}{(3q - 4)(q - 5)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{9q^2 - 12q - (2q^2 - 15q + 25)}{(q - 5)(3q - 4)} $ $ = \dfrac{9q^2 - 12q - 2q^2 + 15q - 25}{(q - 5)(3q - 4)} $ $ = \dfrac{7q^2 + 3q - 25}{(q - 5)(3q - 4)}$ Expand the denominator: $ = \dfrac{7q^2 + 3q - 25}{3q^2 - 19q + 20}$